35^2=b^2+20^2

Solution for 35^2=b^2+20^2 equation:

35^2=b^2+20^2

We move all terms to the left:

35^2-(b^2+20^2)=0

We add all the numbers together, and all the variables

-(b^2+20^2)+1225=0

We get rid of parentheses

-b^2+1225-20^2=0

We add all the numbers together, and all the variables

-1b^2+825=0

a = -1; b = 0; c = +825;
Δ = b2-4ac
Δ = 02-4·(-1)·825
Δ = 3300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3300}=\sqrt{100*33}=\sqrt{100}*\sqrt{33}=10\sqrt{33}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{33}}{2*-1}=\frac{0-10\sqrt{33}}{-2}
=-\frac{10\sqrt{33}}{-2}
=-\frac{5\sqrt{33}}{-1}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{33}}{2*-1}=\frac{0+10\sqrt{33}}{-2}
=\frac{10\sqrt{33}}{-2}
=\frac{5\sqrt{33}}{-1}
$